Why do we require quotient to be surjective? 
Definition. Let $X$ and $Y$ be topological spaces; let $p : X \to Y$ a
  surjective map. The map $p$ is said to be a quotient map provided a
  subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.
Definition. If $X$ is a space and $A$ is a set and if $p : X\to A$ is a surjective map, then there exists exactly one topology $\tau$ on
  $A$ relative to which $p$ is a quotient map; it is called the quotient
topology induced by $p$.
Definition. Let X be a topological space, and let $X^*$ be a partition of $X$ into disjoint subsets whose union is $X$. Let $p : X  \to X^*$ be the surjective map that carries each point of $X$ to the element of $X^*$ containing it. In the quotient topology induced by $p$, the space $X^ *$ is called a quotient space of $X$

It seems that to induce a quotient topology by using $p$, we don't need the property of $p$ being surjective. Why do we require a quotient map to be surjective? Are there some important applications of the surjection property? To prove a function is a quotient map, do we need to explicitly prove the function is surjective ? 
Does the following picture give some intuition? It's from Wikipedia. The next stage of this gif is the $S^2$ obtained by gluing the boundary (in blue) of the disk $D^2$ together to a single point.

 A: The quotient space has a universal property, namely if $f:X\rightarrow Z$ is a continuous function with the property that $(p(a)=p(b))\Rightarrow (f(a)=f(b))$, then $f$ factors through the quotient space.  In other words, there exists a unique continuous function $\widetilde{f}:Y\rightarrow Z$ so that $f=\widetilde{f}\circ p$.
$$
\begin{matrix}
X && \\
{\scriptsize p} \downarrow & \  \searrow^{f} & \\
Y & \underset{\bar f}{\longrightarrow} & Z 
\end{matrix}
$$
If you don't take the quotient map to be surjective, this property fails (the map fails to be unique).
From the definition you state, it seems that one could define a quotient even if map $p$ is not surjective, but this breaks a desired property (that might not have been mentioned in your studies yet).
If you consider a "quotient map" $q$ that has all the properties above, but isn't surjective, then you can write $q$ as a composition of a (true) quotient and an inclusion map.  In other words, the composition $X\stackrel{p}{\rightarrow} Y\stackrel{i}{\hookrightarrow}Z$ equals $q$.
$$
\begin{matrix}
X && \\
{\scriptsize p} \downarrow & \ \searrow^q & \\
Y & \underset{i}{\hookrightarrow} & Z
\end{matrix}
$$
A: If p is not surjective, then for all y not in p(X), {y} is open.
Surjectivity removes the superflous open singles.
It also allows the equivalence relation x ~ y
when p(x) = p(y) to uniquely partition Y.
A: A quotient is a division. In the case of a set, even if it has additional structure like a topological space, this means we divide the set into partitions. Each partition is non-empty and defines an equivalence class, i.e. if $x$ and $y$ are in the same partition they are equivalent.
The simplest equivalence relation you can define is that each element of the set is only equivalent to itself. In that case, the number of equivalence classes, or partitions, is equal to the number of elements of the set and that is the largest number of equivalence classes you can have.
Since a quotient map maps each element to its equivalence class it must be surjective.
